我在PHP中遇到了preg_replace函数的问题。我无法弄清楚模式和替代品。
我有两个字符串和一些代码:
$dirname1 = 'hdadas/dasdad/dasd/period_1min';
$dirname2 = 'hdadas/dasdad/dasd/period_1min/abcdrfg.php';
$pieces1 = explode('/', $dirname1);
$pieces2 = explode('/', $dirname2);
$dirname1 = end($pieces1); // output will be period_1min
$dirname2 = end($pieces2); // output will be abcdrfg.php
$output = preg_replace($pattern, $replacement, $dirname1); // or (..,..,$dirname2
echo $output; // i need 1min(without period_) or abcdrfg (without .php)
UPD:
function Cat($dirname)
{
$name = explode('/', $dirname);
$pattern = ???;
$replacement = ???;
return preg_replace($pattern, $replacement, $dirname1);
}
print(Cat('hdadas/dasdad/dasd/period_1min'))); // output need 1min only
print(Cat('hdadas/dasdad/dasd/period_1min/abcdrfg.php'))); // output need abcdrfg only
答案 0 :(得分:2)
这应该适合你:
(这里我使用basename()
仅获取路径的最后部分,然后使用pathinfo()
获取没有扩展名的最后一部分。之后我只使用preg_replace()
来用空字符串替换下划线之前的所有内容)
<?php
$dirname1 = "hdadas/dasdad/dasd/period_1min";
$dirname2 = "hdadas/dasdad/dasd/period_1min/abcdrfg.php";
$dirname1 = pathinfo(basename($dirname1))["filename"];
$dirname2 = pathinfo(basename($dirname2))["filename"];
echo $output = preg_replace("/(.*)_/", "", $dirname1);
?>
输出:
1min
abcdrfg
答案 1 :(得分:1)
这个正则表达式如何/^.+_|\.[^.]+$/
:
$dirname1 = 'hdadas/dasdad/dasd/period_1min';
$dirname2 = 'hdadas/dasdad/dasd/period_1min/abcdrfg.php';
$pieces1 = explode('/', $dirname1);
$pieces2 = explode('/', $dirname2);
$dirname1 = end($pieces1); // output will be period_1min
$dirname2 = end($pieces2); // output will be abcdrfg.php
$output = preg_replace('/^.+_|\.[^.]+$/', '', $dirname1); // or (..,..,$dirname2
echo $output,"\n"; // i need 1min(without period_) or abcdrfg (without .php)
$output = preg_replace('/^.+_|\.[^.]+$/', '', $dirname2); // or (..,..,$dirname2
echo $output,"\n"; // i need 1min(without period_) or abcdrfg (without .php)
<强>输出:强>
1min
abcdrfg